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The Constants to Measure the Differences Between Isosceles and α− β Orthogonalities

Qichuan Ni, Qi Liu∗, Yin Zhou, Qin Qian
School of Mathematics and Physics, Anqing Normal University, Anqing 246133, P. R. China

niqichuan111@163.com, liuq67@aqnu.edu.cn, zhouyin0330@163.com, 15212956918@163.com
∗Correspondence: liuq67@aqnu.edu.cn

Abstract. In this paper, by combining the isosceles orthogonality and α−β orthogonality of Banachspaces, we first introduce a new geometric constant. We demonstrate some basic properties aboutit, such as calculating its value in the common norm spaces. Moreover, the necessary and sufficientconditions for the new constant to characterize Hilbert spaces are given. Finally, only consider thepoints on the unit sphere, we introduce another new geometric constant and some basic propertiesare also obtained.

1. Introduction
Traditional orthogonality plays a key role in the geometry of Banach spaces and is a geometricfeature of Hilbert spaces. We repeat the definitions of the next three orthogonality types. In 1945,James [8] introduced isosceles orthogonality as follows:

x ⊥I y if and only if ‖x + y‖ = ‖x − y‖.
Balestro [4] introduced the orthogonality of Pythagoras as follows:

x ⊥P y if and only if ‖x − y‖2 = ‖x‖2 + ‖y‖2.
Birkhoff defined the following Birkhoff orthogonality [5] in linear metric spaces:

x ⊥B y if and only if ‖x + αy‖ ≥ ‖x‖ for all α ∈ R.
These three orthogonalities have been investigated in several papers (see [2], [9] and so on).Over the years, many scholars have introduced the concept of extended orthogonality. The familyof orthogonalities introduced by Carlson [6] in 1961, covering isosceles and Pythagorean orthogo-nalities, is known as Carlson orthogonality:

x ⊥C y if and only if n∑
i=1

ai‖bix + ciy‖2 = 0.

Received: 22 Jan 2024.
Key words and phrases. Isosceles orthogonality; α− β orthogonality; Geometric constant; Hilbert spaces.1

https://adac.ee
https://doi.org/10.28924/ada/ma.4.6


Eur. J. Math. Anal. 10.28924/ada/ma.4.6 2A further special case of Carlson orthogonality was introduced by Dimitini et al. [7] in 1983 asfollows:
x ⊥α y if and only if (1 + α2)‖x − y‖2 = ‖x − αy‖2 + ‖y − αx‖2,where fixed α 6= 1. Two years later, this orthogonality was generalized by the same author [3] as:

x ⊥αβ y if and only if ‖x − y‖2 + ‖αx − βy‖2 = ‖x − βy‖2 + ‖y − αx‖2.
It is well known that in the general Banach spaces, these orthogonalities are not the same. Forexample, isosceles orthogonality has symmetry, but Birkhoff orthogonality does not have symmetry,which means that x ⊥B y cannot show that y ⊥B x . Because of this difference, measuring thedifference between the two types of orthogonality is of great significance. Many scholars havedefined and studied many novel orthogonality geometric constants and given a large number ofresults, including well-known constants (see [11], [12]):

BI(X) = sup

{
‖x + y‖ − ‖x − y‖

‖x‖ : x, y ∈ SX , x, y 6= 0, x ⊥B y
}

and
BR(X) = sup

α>0

{
‖x + αy‖ − ‖x − αy‖

α
: x, y ∈ SX , x ⊥B y

}
.

For more information, refer to references [1], [13-15]. Throughout the article, we use X torepresent a real Banach space with norm ‖ · ‖, the unit ball is denoted as BX = {x ∈ X : ‖x‖ ≤ 1}and the unit sphere is denoted as SX = {x ∈ X : ‖x‖ = 1}. Let’s assume that dimX is greaterthan or equal to 2.
2. The Constant Lα,β(X)

Combining the extended isosceles and Pythagorean orthogonalities, we define a new constantto describe the difference between these two orthogonalities.
Definition 2.1. Let X be a Banach space, the geometric constant of the isosceles orthogonality
type is defined as

Lα,β(X) = sup

{
‖x − βy‖2 + ‖αx − y‖2

‖x − y‖2 + ‖αx − βy‖2 : x ⊥I y , (x, y) 6= (0, 0)
}
, where α, β ≥ 0, α 6= 1, β 6= 1.

In this section, we first give bounds on the constant Lα,β(X) and its value in some particularspaces is obtained. We find that X is a Hilbert space if and only if the constant Lα,β(X) value is1. Moreover, the relationship between Lα,β(X) and uniformly non-square is given. Now, we recallthe notion of the uniformly non-square.
Definition 2.2. ([10]) A Banach space X is called uniformly non-square if there exists δ ∈ (0, 1)
such that for any x, y ∈ SX , then

‖x + y‖
2

≤ 1− δ or ‖x − y‖
2

≤ 1− δ.

https://doi.org/10.28924/ada/ma.4.6


Eur. J. Math. Anal. 10.28924/ada/ma.4.6 3

Proposition 2.1. Let X be a Banach space, then

1 ≤ Lα,β(X) ≤



2
1+β2

, 0 ≤ β ≤ α < 1;
2

1+α2
, 0 ≤ α ≤ β < 1;

α2+β2

1+β2
, 1 < β ≤ α;

α2+β2

1+α2
, 1 < α ≤ β.

proof. Let x0 = 0, y0 6= 0, satisfy x0 ⊥I y0, hence
Lα,β(X) ≥

‖x0 − βy0‖2 + ‖αx0 − y0‖2

‖x0 − y0‖2 + ‖αx0 − βy0‖2
= 1.

On the other hand, using the triangle inequality, we get:
‖x − βy‖2 + ‖αx − y‖2

‖x − y‖2 + ‖αx − βy‖2

≤

( ∣∣∣1−β2 ∣∣∣ ‖x + y‖+ ∣∣∣1+β2 ∣∣∣ ‖x − y‖)2 + ( ∣∣α−12 ∣∣ ‖x + y‖+ ∣∣α+12 ∣∣ ‖x − y‖)2
‖x − y‖2 +

( ∣∣∣α−β2 ∣∣∣ ‖x + y‖ − ∣∣∣α+β2 ∣∣∣ ‖x − y‖)2 .

If we take the categorical approach, there will be the following four approaches.
Case 1 When 0 ≤ β ≤ α < 1, we have

‖x − βy‖2 + ‖αx − y‖2

‖x − y‖2 + ‖αx − βy‖2 ≤
2

1 + β2
.

Case 2 When 0 ≤ α ≤ β < 1, we have
‖x − βy‖2 + ‖αx − y‖2

‖x − y‖2 + ‖αx − βy‖2 ≤
2

1 + α2
.

Case 3 When 1 < β ≤ α, we have
‖x − βy‖2 + ‖αx − y‖2

‖x − y‖2 + ‖αx − βy‖2 ≤
α2 + β2

1 + β2
.

Case 4 When 1 < α ≤ β, we have
‖x − βy‖2 + ‖αx − y‖2

‖x − y‖2 + ‖αx − βy‖2 ≤
α2 + β2

1 + α2
.

�

Next, we will show that there are points in some particular spaces where the value of the constant
Lα,β(X) is an upper bound.
Example 2.1. Let 0 ≤ β < 1, α = β and X be the space R2 with l1 norm defined by

‖(x1, x2)‖ = |x1|+ |x2|,

then Lα,β(X) = 2
1+β2

.

https://doi.org/10.28924/ada/ma.4.6


Eur. J. Math. Anal. 10.28924/ada/ma.4.6 4

Let x = (1, 1), y = (1,−1), satisfy x ⊥I y . We get ‖αx − βy‖ = 2β, ‖x − y‖ = ‖x − βy‖ =
‖αx − y‖ = 2.

Thus, Lα,β(X) = 2
1+β2

.

Example 2.2. Let 0 ≤ β < 1, α = β and X be the space R2 with l∞ norm defined by

‖(x1, x2)‖ = max{|x1|, |x2|},

then Lα,β(X) = 2
1+β2

.
Let x = (1, 0), y = (0,−1), satisfy x ⊥I y . We get ‖αx − βy‖ = β, ‖x − y‖ = ‖x − βy‖ =

‖αx − y‖ = 1.
Thus, Lα,β(X) = 2

1+β2
.

Proposition 2.2. Let X be a Banach space, 0 ≤ β ≤ α < 1, then Lα,β(X) = 1 if and only if X is
a Hilbert space.

proof. Since X is a Hilbert space, combined with reference [3], we have Lα,β(X) = 1. Conversely,assuming that Lα,β(X) = 1 and useing the homogeneity of x ⊥ y , we can prove by induction that
‖x − y‖2 + ‖αnx − βy‖2 ≥ ‖αnx − y‖2 + ‖x − βy‖2.

Since 0 ≤ α < 1, taking the limit n →∞, we get
‖x − y‖2 ≥

(
1− β2

)
‖y‖2 + ‖x − βy‖2.

If 0 ≤ β < 1, a second induction shows that ‖x − y‖2 = (1− β2n) ‖y‖2+ ‖x − βny‖2. Also takingthe limit n →∞, Therefore, in this case we get ‖x − y‖2 ≥ ‖x‖2 + ‖y‖2. Then, x ⊥I y implies
‖x + y‖2 + ‖x − y‖2 ≥ 2‖x‖2 + 2‖y‖2

for all x, y ∈ X , hence we can assert that X is a Hilbert space. �

Among the many properties of Banach spaces, we give below a sufficient condition that X is nota uniform non-square space. In the process of proving, we apply the lemma given by James.
Lemma 2.1. [8, Lemma 4.1] Let X be a Banach space and x, y ∈ X . If x ⊥I y , then the following
inequality holds.

(i) ‖x + ky‖ ≤ |k |‖x ± y‖ and ‖x ± y‖ ≤ ‖x + ky‖, when |k | ≥ 1.
(ii) ‖x + ky‖ ≤ ‖x ± y‖ and |k |‖x ± y‖ ≤ ‖x + ky‖, when |k | ≤ 1.

Proposition 2.3. Let X be a finite dimensional Banach space, if Lα,β(X) = 2
1+β2

for some 0 ≤
β0 ≤ α0 < 1, then X is not uniformly non-square.

proof. Since Lα,β(X) = 2
1+β2

, there exist xn ∈ SX , yn ∈ BX that satisfy xn ⊥I yn and
lim
n→∞

‖xn − βyn‖2 + ‖αxn − yn‖2

‖xn − yn‖2 + ‖αxn − βyn‖2
=

2

1 + β2
.

https://doi.org/10.28924/ada/ma.4.6


Eur. J. Math. Anal. 10.28924/ada/ma.4.6 5At the same time, a Banach space X is finite dimensional, then there exist x0, y0 ∈ BX that satisfy
x0 ⊥I y0 and

lim
k→∞

∥∥xnk∥∥ = ‖x0‖ , lim
k→∞

∥∥ynk∥∥ = ‖y0‖ .
Combine lemma 2.1, we have ‖xn − β0yn‖ ≤ ‖xn + yn‖ and ‖α0xn − yn‖ ≤ ‖xn + yn‖.Thus,

‖xn + yn‖2 + ‖xn + yn‖2

‖xn + yn‖2 + (
∣∣∣α−β2 ∣∣∣− ∣∣∣α+β2 ∣∣∣)2‖xn + yn‖2 ≤

2

1 + β2
,

‖xn + yn‖2 + ‖xn + yn‖2

(1 + β2) ‖xn + yn‖2
≤

2

1 + β2
.

We can obtain ‖x0 − β0y0‖ = ‖x0 + y0‖ and ‖α0x0 − y0‖ = ‖x0 + y0‖. Since ‖x0 − β0y0‖ ≤
(1− β0) ‖x0‖+β0 ‖x0 + y0‖, then ‖x0 + y0‖ ≤ ‖x0‖. Moreover, we can prove that ‖x0 + y0‖ ≤ ‖y0‖,then

max {‖x0 + y0‖ , ‖x0 − y0‖} = ‖x0 + y0‖ ≤ min {‖x0‖ , ‖y0‖} ≤ 1 < 1 + δ

for any δ ∈ (0, 1), this means that X is not uniformly non-square. �

3. The Constant L′α,β(X)In this section, if x and y satisfy the isosceles orthogonality condition and restrict x , y ∈ SX ,then we define the new constant:
Definition 3.1. Let X be a Banach space, another geometric constant of the isosceles orthogonal
type is defined as

L′α,β(X) = sup

{
‖x − βy‖2 + ‖αx − y‖2

‖x − y‖2 + ‖αx − βy‖2 : x, y ∈ SX , x ⊥I y
}
, where α, β ≥ 0, α 6= 1, β 6= 1.

We give bounds on the constant L′α,β(X) and calculate its value in `∞− `1 normed linear spacewhen α = β = 1
2 .

Proposition 3.1. Let X be a Banach space, then

α2+β2

1+α2

α2+β2

1+β2

2
1+α2

2
1+β2

≤ L′α,β(X) ≤



2
1+β2

, 0 ≤ β ≤ α < 1;
2

1+α2
, 0 ≤ α ≤ β < 1;

α2+β2

1+β2
, 1 < β ≤ α;

α2+β2

1+α2
, 1 < α ≤ β.

https://doi.org/10.28924/ada/ma.4.6


Eur. J. Math. Anal. 10.28924/ada/ma.4.6 6

proof. Combined with the idea of Proposition 2.1, we have
‖x − βy‖2 + ‖αx − y‖2

‖x − y‖2 + ‖αx − βy‖2

≥

( ∣∣∣1−β2 ∣∣∣ ‖x + y‖ − ∣∣∣1+β2 ∣∣∣ ‖x − y‖)2 + ( ∣∣α−12 ∣∣ ‖x + y‖ − ∣∣α+12 ∣∣ ‖x − y‖)2
‖x − y‖2 +

( ∣∣∣α−β2 ∣∣∣ ‖x + y‖+ ∣∣∣α+β2 ∣∣∣ ‖x − y‖)2 .

Case 1 When 0 ≤ β ≤ α < 1, we have
‖x − βy‖2 + ‖αx − y‖2

‖x − y‖2 + ‖αx − βy‖2 ≥
α2 + β2

1 + α2
.

Case 2 When 0 ≤ α ≤ β < 1, we have
‖x − βy‖2 + ‖αx − y‖2

‖x − y‖2 + ‖αx − βy‖2 ≥
α2 + β2

1 + β2
.

Case 3 When 1 < β ≤ α, we have
‖x − βy‖2 + ‖αx − y‖2

‖x − y‖2 + ‖αx − βy‖2 ≥
2

1 + α2
.

Case 4 When 1 < α ≤ β, we have
‖x − βy‖2 + ‖αx − y‖2

‖x − y‖2 + ‖αx − βy‖2 ≥
2

1 + β2
.

On the other hand, the upper bound on the constant L′α,β(X) is the same as the upper bound onthe constant Lα,β(X). �

Example 3.1. Let α = β = 1
2 and X be the space R2 with `∞ − `1 norm defined by

‖x‖ =

{
‖x‖1, x1x2 ≤ 0,
‖x‖∞, x1x2 ≥ 0.

Then

L′1
2
, 1
2

(X) = 0.91.

proof. If x = (y1, 1+y1), y = (y2, 1+y2), where −1 ≤ y1 ≤ y2 ≤ 0; x = (y1, y1−1), y = (y2, y2−1),where 0 ≤ y1 ≤ y2 ≤ 1. The two cases above, which are determined by x ⊥I y , we have
|y1 − y2| = 2, are contradictory. To estimate this constant value, it is only necessary to considerthe following two cases.

Case 1: Assuming that x = (x1, 1) , y = (1, y2), 0 ≤ x1 ≤ y2 ≤ 1. Since x ⊥I y , we have
1 + y2 = (1− x1) + (1− y2) ,

https://doi.org/10.28924/ada/ma.4.6


Eur. J. Math. Anal. 10.28924/ada/ma.4.6 7hence x1 + 2y2 = 1, y2 ∈ [13 , 12]. Then ‖αx − y‖ = 1, ‖x − βy‖ = 1
2 +

3
2y2, ‖αx − βy‖ = 1

2 +
1
2y2and ‖x − y‖ = 1 + y2.In fact,

L′1
2
, 1
2

(X) = max
1
3
≤y2≤ 12

9y22 + 6y2 + 5

5y22 + 10y2 + 5
.

By simple calculation, we find that L′1
2
, 1
2

(X) = 0.91 is obtained at the point (1, 12).
Case 2: Assuming that x = (x1, 1) , y = (y1, 1 + y1) satisfy −1 ≤ y1 ≤ 0 ≤ x1 ≤ 1. Since

x ⊥I y , we have ‖(x1 + y1, 2 + y1)‖ = ‖(x1 − y1,−y1)‖.If −x1 ≤ y1, then 2 + y1 = x1 − y1 is true, hence x1 − 2y1 = 2, y1 ∈ [−23 ,−12] . We have
‖αx − y‖ = 1, ‖x − βy‖ = 2 + 32y1, ‖αx − βy‖ = 1 + 12y1 and ‖x − y‖ = 2 + y1.In the same way,

L′1
2
, 1
2

(X) = max
− 2
3
≤y1≤− 12

9y21 + 24y1 + 20

5y21 + 20y1 + 20
.

By simple calculation, we find that L′1
2
, 1
2

(X) = 0.91 is obtained at the point (1,−12).Similarly, if y1 ≤ −x1, such as Case 2, prove omission.Combined with all of the above, we get
L′1
2
, 1
2

(X) = 0.91.

�

4. Funding Statement
This research work was funded by Anhui Province Higher Education Science Research Project(Natural Science), 2023AH050487.

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	1. Introduction
	2. The Constant L,(X)
	3. The Constant L,(X)
	4. Funding Statement
	References

